Chemistry, 27.09.2019 16:30 therealpr1metime45

Consider this reaction and answer the questions:

kcio3-> kcl+o2

what is the oxidation number of chlorine in the reactant state?
what is the oxidation number of oxygen in the reactant state?
what is the oxidation number of potassium in the reactant state?
what is the oxidation number of chlorine in the product state?
what is the oxidation number of oxygen in the product state?
what is the oxidation number of potassium in the product state?

Answers: 1

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Answers

Answer from: Wintersavannah

1) +5

2) -2

3) +1

4) -1

5) 0

6) +1

Explanation:

The rules to determine oxidation number are

a) The oxidation number of alkali metals is +1 (except in super oxides or peroxides).

b) the oxidation number of oxygen is -2 (except in super and peroxide or compounds with fluorine or with itself)

1) the oxidation number of chlorine in the reactant state: KClO₃:

+1 + x + 3(-2) = 0

x = oxidation number of chlorine =+5

2) The oxidation number of oxygen in the reactant state: -2 as it is a normal oxide.

3) The oxidation number of potassium in the reactant state: +1

4) The oxidation number of chlorine in the product state: -1

5 The oxidation number of oxygen in the product state: 0

The compound with itself.

6) The oxidation number of potassium in the product state+ +1 (alkali metal)

Answer from: awuekure

2KCl⁺⁵O₃ → 2KCl + 3O₂⁰

reagents
Cl +5
O -2
K +1

products
Cl 0
O -2
K +1

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.

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kcio3-> kcl+o2

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